1. Choose the rotating frame of the table

   The ball moves inside a smooth radial groove, so relative to the table it can move only along the radial direction.

   Let the radial speed of the ball relative to the table be $\dot r$.

2. Forces in the rotating frame

   In the frame rotating with angular velocity $\omega$, the ball experiences:

   • Centrifugal force outward: $m\omega^2 r$
   • Coriolis force: $2m\omega \dot r$, perpendicular to the groove

   Since the groove is radial, the Coriolis force is balanced by the side reaction of the groove and does not affect radial motion.

   Hence radial equation of motion is $m\ddot r = m\omega^2 r$ or $\ddot r = \omega^2 r.$

3. Use energy-type integration

   Multiply by $\dot r$: $\ddot r\,\dot r = \omega^2 r\dot r.$

   Integrating, $\frac{1}{2}\dot r^2 = \frac{1}{2}\omega^2 r^2 + C.$

4. Initial conditions

   The ball is gently placed at $r=1\,\text{m}$, so initially relative to the table, $\dot r = 0 \quad \text{at} \quad r=1.$

   Therefore, $0 = \omega^2(1)^2 + 2C \quad \Rightarrow \quad C=-\frac{1}{2}\omega^2.$

   So, $\dot r^2 = \omega^2(r^2-1).$

5. At the instant the ball leaves the table

   The table radius is $3\,\text{m}$, so the ball leaves when $r=3.$

   Then $\dot r^2 = \omega^2(9-1)=8\omega^2.$

   Hence $\dot r = 2\sqrt{2}\,\omega.$

6. Compare with the given form

   Given radial velocity is $x\sqrt{2}\,\omega.$

   Therefore, $x=2.$

Comparison with stored answer: Stored correct answer is $2$, which matches our result.